【xsy1504】 pitcure 树状数组

数据范围：$n≤2 imes 10^5$

 

以下是题解：

#include<bits/stdc++.h>
#define L unsigned int
#define MOD 16777216
#define lowbit(x) ((x)&(-(x)))
#define M 200005
using namespace std;

L b[M]={0},a[M]={0},n;
void add(L x,L k){for(L i=x;i<=n;i+=lowbit(i)) b[i]+=k;}
L query(L x){L k=0; for(L i=x;i;i-=lowbit(i)) k+=b[i]; return k;}

L solve1(){
    static L r[M]={0},j[M]={0}; L res=0;
    memset(b,0,sizeof(b));
    for(L i=n;i;i--){
        r[i]=query(n)-query(a[i]);
        add(a[i],1);
    }
    memset(b,0,sizeof(b));
    for(L i=1;i<=n;i++){
        j[i]=query(a[i]);
        add(a[i],i);
    }
    memset(b,0,sizeof(b));
    for(L i=1;i<=n;i++){
        L l=query(a[i]),now=0;
        now+=1LL*l*(i-1);
        now-=1LL*l*(l-1)/2;
        now-=j[i];
        res+=1LL*now*r[i];
        add(a[i],1);
    }
    return res;
}
L solve2(){
    static L r[M]={0},j[M]={0}; L res=0;
    memset(b,0,sizeof(b));
    for(L i=n;i;i--){
        r[i]=query(n)-query(a[i]);
        add(a[i],1);
    }
    memset(b,0,sizeof(b));
    for(L i=1;i<=n;i++){
        j[i]=query(a[i]);
        add(a[i],a[i]);
    }
    memset(b,0,sizeof(b));
    for(L i=1;i<=n;i++){
        L l=query(a[i]),now=0;
        now+=1LL*l*(a[i]-1);
        now-=1LL*l*(l-1)/2;
        now-=j[i];
        res+=1LL*now*r[i];
        add(a[i],1);
    }
    return res;
}
    
L solve3(){
    L res=0; memset(b,0,sizeof(b));
    for(L i=n;i;i--){
        L hh=query(n)-query(a[i]);
        res+=(1LL*hh*(hh-1)*(hh-2)/6);
        add(a[i],1);
    }
    return res;
}
L solve4(){
    static L f[M]={0};
    for(L i=n;i;i--) f[i]=1;
    for(L x=1;x<=3;x++){
        memset(b,0,sizeof(b));
        for(L i=n;i;i--){
            L hh=query(n)-query(a[i]);
            add(a[i],f[i]);
            f[i]=hh;
        }
    }
    L res=0; for(L i=1;i<=n;i++) res+=f[i];
    return res;
}

main(){
    scanf("%u",&n);
    for(L i=1;i<=n;i++) scanf("%u",a+i);
    L res1=solve1();//1x2x
    L res2=solve2();//14xx
    L res3=solve3();//1xxx
    L res4=solve4();//1234
    printf("%u
",((res1+res2-res3+res4)%MOD+MOD)%MOD);
}